Math, asked by unique1man, 7 months ago

prove that1/2(sin7A-sin3A)=sin5Asin2A*sin5A

Answers

Answered by Anonymous

7

$\huge \mathbf{solution}$

sinA+sin3A+sin5A+sin7A/cosA+cos3A+cos5A+cos7A

=(sinA+sin7A)+(sin3A+sin5A)/(cosA+cos7A)+(cos3A+cos5A)

=(2sin4Acos3A+2sin4AcosA)/(2cos4Acos3A+2cos4AcosA)

=2sin4A(cos3A+cosA)/2cos4A(cos3A+cosA)

=sin4A/cos4A

=tan4A Hence it is proved

Answered by Anonymous

16

see in attachment for answer ✌️

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